Problem: Factor the following expression: $x^2 + 13x + 40$
Answer: When we factor a polynomial, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + a)(x + b) &=& xx &+& xb + ax &+& ab \\ \\ &=& x^2 &+& {(a + b)}x &+& {ab} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + a)(x + b) }&\hphantom{=}&\hphantom{ xx }&\hphantom{+}&\hphantom{ (a + b)x }&\hphantom{+}& \\ &=& x^2 & +& {13}x& +& {40} \end{eqnarray} $ The coefficient on the $x$ term is $13$ and the constant term is $40$ , so to reverse the steps above, we need to find two numbers that add up to $13$ and multiply to $40$ You can try out different factors of $40$ to see if you can find two that satisfy both conditions. If you're stuck and can't think of any, you can also rewrite the conditions as a system of equations and try solving for $a$ and $b$ $ {a} + {b} = {13}$ $ {a} \times {b} = {40}$ The two numbers $5$ and $8$ satisfy both conditions: $ {5} + {8} = {13} $ $ {5} \times {8} = {40} $ So we can factor the expression as: $(x + {5})(x + {8})$